Question

a bacterial culture is known to grow at a rate proportional to the amt. present.after 1 hour, 1000 strands of the bacteria are observed in the culture, and after 4 hours, 3000 strands. find the number of strands of bacteria originally in the culture

Answer 1

at t= 0
x = not given

at t = 1
x = 1000

at t = 4
x = 3000

Answer 2

\[\ln(\frac{3000}{1000} = k(4-1)\]
\[\ln (3) = 3k\]
\[k = 0.3662\]

Answer 3

\[\huge{\frac{dA}{dt}=kA}\]
where A is the Amount of bacteria and k is a constant.

Solve the differential equation and substitute the values to get k and the constant of integration.

then put t = 0

Answer 4

Source : http://www.mathisfunforum.com/viewtopic.php?id=17940

Answer 5

\[x = x_o e^{0.3672t}\]
\[1000 = x_0 e^{0.3662(1)}\]
\[1000 = 1.4422x_0\]
\[x_0 = 693.36 \implies 694 \]

Answer 6

you know it triples every 3 hours, so can model as
\[1000\times 3^\frac{t}{3}\] where \(t\) is the time starting at hour one
to get one hour previous, replace \(t\) by -1

Answer 7

yikes no need for \(e\) and all that fancy jazz

Answer 8

no need for differential equations either
you are told directly that it triples every three hours, use that immediately

Answer 9

how would you solve it w/o d.e.?

Answer 10

i just solved it, you can do everything but the last calculation
http://www.wolframalpha.com/input/?i=1000*3^%28-1%2F3%29
in your head

Answer 11

wait why \(\large 1000 \times 3^{\frac t3}?\) i didnt quite get your explanation for it sorry

Answer 12

your are told the growth rate is proportional in the first line
in the second line you are told that it triples every three hours

Answer 13

mmhmm

Answer 14

wait..t = 1... x = 1000
t = 3...x = 3000

isnt it per 2 hours?

Answer 15

so every three hours it triples right?
if you start counting at \(t=0\) you have \(p_0\)
at \(t=3\) you have \(p_0\times 3\)
at \(t=6\) you have \(p_0\times 3^2\)
at \(t=9\) you have \(p_0\times 3^3\) and more generally
at \(t=t\) you have \(p_0\times 3^{\frac{t}{3}}\)

Answer 16

you started counting at hour 1, not hour zero

Answer 17

ohh im seeing your logic now

Answer 18

that is why i had to replace \(t\) by \(-1\) to get back one step

Answer 19

hmm son of a gun..it's right...

Answer 20

this solving for the rate is a waste of time as far as i am concerned because you are told the rate in the problem
triples every three hours

Answer 21

you raise a good point...but it's some out of the box thinking to come up with a relation like that..

Answer 22

population increases from 100 to 150 every two hours
model immediately as \(100\times (1.5)^{\frac{t}{2}}\) no thought required

Answer 23

no no it is simple thinking
keeps me from having to use a different base, when in fact you are told the base
the truth is you can use \(e\), \(3\) or \(\pi\) if you like it is just that if you are told the base in the question you do not need to change it to some other base

Answer 24

i see...i'll try that thanks

Answer 25

so im gonna use the difference as base..?

Answer 26

well your teacher might not like it, but believe me it is fast, works, and is in fact more accurate because you do not truncate as you do when you take the log

Answer 27

what i mean is this:
suppose i am told that the population starts at say 100 and in 2 hours it is 150
that means it increases by 50% every two hours. to increase something by 50% multiply by \(1.5\) which is of course the same as computing \(\frac{150}{100}\)
so now i have my base as \(1.5\) and since it increases by 50% every 2 hours i write
\[100\times (1.5)^{\frac{t}{2}}\]
works the same for half life

Answer 28

ohhh i think im beginning to get the idea

Answer 29

whereas the other method requires me to solve
\[100e^{2k}=150\] for \(k\) and then go back to write my formula

Answer 30

hmm i agree that diff eq is long..

Answer 31

my car ( a nice old satellite) depreciates from 1200 to 800 in 3 years
i model immediately as \(1200\times (\frac{2}{3})^{\frac{t}{3}}\)